If x is an integer, and
x + (x + 1) + … + 2020 + 2021 = 2021
then of course
x + (x + 1) + … + 2020 = 0
and we can write the left side as a sum,
[tex]\displaystyle \sum_{k=0}^n (x+k) = 0[/tex]
with n such that x + n = 2020. (Notice that there are n + 1 terms in the sum.) Then x = 2020 - n, so that
[tex]\displaystyle \sum_{k=0}^n (2020 - n + k) = (2020 - n) \sum_{k=0}^n 1 + \sum_{k=0}^n k= 0[/tex]
Recall that
[tex]\displaystyle \sum_{k=1}^n 1 = 1 + 1 + 1 + \cdots + 1 = n[/tex]
[tex]\displaystyle \sum_{k=1}^n 1 = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2[/tex]
Then
[tex]\displaystyle (2020 - n) (n + 1) + \frac{n(n+1)}2= 0[/tex]
Solve for n :
(2020 - n) (n + 1) + n (n + 1) / 2 = 0
4040 + 4039 n - n² = 0
(n - 4040) (n + 1) = 0
⇒ n = 4040 or n = -1
But we are taking the sum of a positive number of terms, so n = 4040.
Then since x + n = 2020, it follows that x = 2020 - 4040 = -2020.
